If `x^a=y^b=c^c`, where `a,b,c` are unequal positive numbers and `x,y,z` are in GP, then `a^3+c^3` is :

Options

A. `ge 2b^3`
B. `gt 2b^3`
C. `le 2b^3`
D. `lt 2b^3`

Answer

To solve the problem, we are given that \( x^a = y^b = z^c \), where \( a, b, c \) are unequal positive numbers and \( x, y, z \) are in geometric progression (GP). We need to find \( a^3 + c^3 \).

### Step-by-Step Solution:

1. **Set a Common Value**:
Let \( k = x^a = y^b = z^c \). Therefore, we can express \( x, y, z \) in terms of \( k \):
\[ x = k^{1/a}, \quad y = k^{1/b}, \quad z = k^{1/c} \]

2. **Use the GP Property**:
Since \( x, y, z \) are in GP, we have:
\[ y^2 = xz \]
Substituting the expressions for \( x, y, z \):
\[ (k^{1/b})^2 = k^{1/a} \cdot k^{1/c} \]
This simplifies to:
\[ k^{2/b} = k^{1/a + 1/c} \]

3. **Equate the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]

4. **Find a Relation**:
Rearranging gives:
\[ \frac{2}{b} = \frac{c + a}{ac} \]
Cross-multiplying yields:
\[ 2ac = b(a + c) \]

5. **Rearranging the Equation**:
From the equation \( 2ac = b(a + c) \), we can express \( a + c \):
\[ a + c = \frac{2ac}{b} \]

6. **Use the Identity for Cubes**:
We know that:
\[ a^3 + c^3 = (a + c)(a^2 - ac + c^2) \]
We can express \( a^2 - ac + c^2 \) using the identity:
\[ a^2 + c^2 = (a + c)^2 - 2ac \]
Thus:
\[ a^3 + c^3 = (a + c)((a + c)^2 - 3ac) \]

7. **Substituting \( a + c \)**:
Substitute \( a + c = \frac{2ac}{b} \):
\[ a^3 + c^3 = \left(\frac{2ac}{b}\right)\left(\left(\frac{2ac}{b}\right)^2 - 3ac\right) \]

8. **Simplification**:
This expression can be further simplified, but we can also analyze the inequality derived from the properties of \( a, b, c \) being in HP:
\[ b < 2\sqrt{ac} \]
This leads us to conclude that:
\[ a^3 + c^3 \geq 2b^3 \]

### Conclusion:
Thus, the final answer we derive from the relationships and properties of the sequences is:
\[ \boxed{2b^3} \]